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2mx integrasjon
Posted: 13/05-2007 17:19
by John Cena54
Fikk disse oppgaven på prøven, vil vite hvordan man løser disse oppgavene
a) Vis at [symbol:integral] (1/xlnx)dx= ln(lnx)+C
b) Vis at [symbol:integral] (lnx/ [symbol:rot] x)dx=2 [symbol:rot] x(lnx-2)+C
Håper noen greier disse oppgavene, takk

Posted: 13/05-2007 18:10
by zell
a)
[tex]\int \frac{1}{x\ln{x}} \rm{d}x = \int \frac{1}{x} \ \cdot \ \frac{1}{\ln{x}} \rm{d}x[/tex]
[tex]u = \ln{x} \ , \ u^, = \frac{1}{x}[/tex]
[tex]\frac{\rm{d}u}{\rm{d}x} = \frac{1}{x} \ \Rightarrow \ \rm{d}x = x\rm{d}u[/tex]
[tex]\int \frac{1}{\cancel{x}} \ \cdot \ \frac{1}{u} \cancel{x}\rm{d}u[/tex]
[tex]\int \frac{1}{u}\rm{d}u = \ln{u} + C[/tex]
[tex]\int \frac{1}{x\ln{x}} \rm{d}x = \ln{(\ln{x})} + C[/tex]
b)
[tex]I = \int \frac{\ln{x}}{\sqrt{x}}\rm{d}x = \int \ln{x} \ \cdot \ x^{-\frac{1}{2}}\rm{d}x[/tex]
Delvis integrasjon.
[tex]v = \ln{x} \ , \ v^, = \frac{1}{x} \ , \ u^, = x^{-\frac{1}{2}} \ , \ u = 2x^{\frac{1}{2}}[/tex]
[tex]I = \ln{x} \ \cdot \ 2x^{\frac{1}{2}} - \int 2x^{\frac{1}{2}} \ \cdot \ \frac{1}{x} \rm{d}x[/tex]
[tex]I = \ln{x} \ \cdot \ 2x^{\frac{1}{2}} - 2\int x^{-\frac{1}{2}}[/tex]
[tex]I = \ln{x} \ \cdot \ 2\sqrt{x} - 4\sqrt{x} + C = 2\sqrt{x}(\ln{x} - 2) + C[/tex]